2008 AMC 10A Problem 18

Question

A right triangle has perimeter

32

and area

20.

What is the length of its hypotenuse?

\dfrac{57}{4}

\dfrac{59}{4}

\dfrac{61}{4}

\dfrac{63}{4}

\dfrac{65}{4}

Let the legs be

y, z

and the hypotenuse

x,

and record perimeter, area, and Pythagorean relations

Square

y + z = 32 - x

and use

y^2 + z^2 = x^2

with

yz = 40

Let the legs be

y, z

and the hypotenuse

x.

Then

y^2 + z^2 = x^2,

y + z = 32 - x,

and

yz = 40.

Squaring the second equation,

\begin{aligned} &(32 - x)^2 \\ &\quad = y^2 + z^2 + 2yz = x^2 + 80. \end{aligned}

This gives

1024 - 64x = 80,

so

x = \dfrac{59}{4}.

Thus, the correct answer is B.

Problem 18 in other years